Answer :


Ideas required to solve the problem:


The general solution of any trigonometric equation is given as –


• sin x = sin y, implies x = nπ + (– 1)ny, where n Z.


• cos x = cos y, implies x = 2nπ ± y, where n Z.


• tan x = tan y, implies x = nπ + y, where n Z.


given,


sec x cos 5x + 1 = 0, 0 < x < π/2


sec x cos 5x = -1


cos 5x = - cos x


- cos x = cos (π – x)


cos 5x = cos (π – x)


If cos x = cos y, implies 2nπ ± y, where n Z.


5x = 2nπ ± (π – x)


5x = 2nπ + (π – x) or 5x = 2nπ – (π – x)


6x = (2n+1)π or 4x = (2n-1)π



But, 0 < x < π/2



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